by : Tính tổng $S=C_{n}^{0}+\frac{{{3}^{2}}-1}{2}C_{n}^{1}+…+\frac{{{3}^{n+1}}-1}{n+1}C_{n}^{n}$
C. $S=\frac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}$
B. $S=\frac{{{4}^{n+1}}+{{2}^{n+1}}}{n+1}-1$
C. $S=\frac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}+1$
D. $S=\frac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}-1$
Hướng dẫn
Chọn D
Ta có $S={{S}_{1}}-{{S}_{2}}$, trong đó
${{S}_{1}}=C_{n}^{0}+\frac{{{3}^{2}}}{2}C_{n}^{1}+\frac{{{3}^{3}}}{3}C_{n}^{2}+…+\frac{{{3}^{n+1}}}{n+1}C_{n}^{n}$
${{S}_{2}}=\frac{1}{2}C_{n}^{1}+\frac{1}{3}C_{n}^{2}+…+\frac{1}{n+1}C_{n}^{n}$
Ta có ${{S}_{2}}=\frac{{{2}^{n+1}}-1}{n+1}-1$
Tính ${{S}_{1}}=?$
Ta có: $\frac{{{3}^{k+1}}}{k+1}C_{n}^{k}={{3}^{k+1}}\frac{n!}{(k+1)!(n-k)!}$ $=\frac{{{3}^{k+1}}}{n+1}\frac{(n+1)!}{(k+1)!\text{ }\!\![\!\!\text{ }(n+1)-(k+1)\text{ }\!\!]\!\!\text{ !}}$$=\frac{{{3}^{k+1}}}{n+1}C_{n+1}^{k+1}$
$\Rightarrow {{S}_{1}}=\frac{1}{n+1}\sum\limits_{k=0}^{n}{{{3}^{k+1}}C_{n+2}^{k+1}}-2C_{n}^{0}$$=\frac{1}{n+1}\left( \sum\limits_{k=0}^{n+1}{{{3}^{k}}C_{n+1}^{k}}-C_{n}^{0} \right)-2C_{n}^{0}$$=\frac{{{4}^{n+1}}-1}{n+1}-2$.
Vậy $S=\frac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}-1$.
